OFFSET
1,2
COMMENTS
Conjecture: T(n,k) = n*k if and only if k = 1 or gcd(n,k) > 1.
Proof from Tamás Fülöp, Nov 07 2025: (Start)
T(n, 1) = n*1 trivially.
(=>)
Suppose T(n, k) = n*k, k != 1, and gcd(n, k) = 1,
then 1 <= a <= k, 1 <= b <= n : gcd(a, b) = 1 and a*n+b*k = n*k.
gcd(n, k) = 1 and b*k = n*(k-a) imply k|(k-a), k|(k-a) implies k|a,
k|a and 1 <= a <= k imply a = k, but then b = n*(k-a)/k = 0, contradicting b >= 1.
(<=)
Let d = gcd(n, k) > 1, n = d*u, k = d*v with gcd(u, v) = 1.
Choose 1 <= t <= d-1 using the Chinese remainder theorem
so that t !== d (mod p) for every prime p|v, gcd(t, u) = 1.
Let x = v*t, then n*x/k = u*t implies n*x == 0 (mod k).
k-x = d*v-v*t = v*(d-t), then gcd(k-x, floor(n*x/k)) = gcd(v*(d-t), u*t) = 1.
Using the second formula with the conditions satisfied, we get T(n, k) = n*k. (End)
LINKS
Tamás Fülöp, Rows n = 1..100 of triangle, flattened
FORMULA
From Tamás Fülöp, Nov 05 2025: (Start)
T(n, k) = max{a*n+b*k : 1 <= a <= k, 1 <= b <= n, a*n+b*k <= n*k, gcd(a, b) = 1}.
T(n, k) = n*k - min{(n*x) mod k : 0 <= x <= k-1, gcd(k-x, floor(n*x/k)) = 1}. (End)
EXAMPLE
T(8, 5) = 38 = 1*8+6*5, where d = gcd(1, 6) = 1 and the loop visits (0,0) d+1 times. - Tamás Fülöp, Nov 05 2025
Table begins:
1;
2, 4;
3, 5, 9;
4, 8, 11, 16;
5, 9, 14, 19, 25;
6, 12, 18, 24, 29, 36;
PROG
(Python)
from math import gcd
def T(n, k): return n*k-min((n*x)%k for x in range(k) if gcd(k-x, n*x//k)==1)
print([T(n, k) for n in range(1, 12) for k in range(1, n+1)]) # Tamás Fülöp, Nov 05 2025
CROSSREFS
KEYWORD
AUTHOR
Peter Kagey, Mar 09 2019
EXTENSIONS
More terms from Tamás Fülöp, Nov 05 2025
STATUS
approved
